💡First, I recognize that the average of 5 numbers is 27. To find the total sum of these numbers, I multiply the average by the number of numbers: 5 times 27 equals 135.Next, when one number is excluded, the average of the remaining 4 numbers becomes 25. To find the new total sum, I multiply the new average by the number of remaining numbers: 4 times 25 equals 100.To determine the excluded number, I subtract the new total sum from the original total sum: 135 minus 100 equals 35.Therefore, the excluded number is 35.

💡Alright, so I have this problem here that asks which of the given functions is both an odd function and increasing on the interval (0, +∞). The options are:A: y = x²B: y = (x - 1)/xC: y = x + 1/xD: y = x - 1/xOkay, let me start by recalling what an odd function is. An odd function satisfies the condition f(-x) = -f(x) for all x in its domain. So, I need to check each function to see if it meets this condition.Starting with option A: y = x².I know that for y = x², f(-x) = (-x)² = x², which is equal to f(x). This means that y = x² is an even function, not an odd function. So, option A is out.Moving on to option B: y = (x - 1)/x.Let me simplify this function first. y = (x - 1)/x can be rewritten as y = 1 - 1/x. Now, let's check if it's odd. f(-x) = 1 - 1/(-x) = 1 + 1/x. On the other hand, -f(x) = -(1 - 1/x) = -1 + 1/x. These two expressions, f(-x) and -f(x), are not equal because 1 + 1/x is not the same as -1 + 1/x. Therefore, option B is not an odd function.Next, option C: y = x + 1/x.Let's check if this is odd. f(-x) = -x + 1/(-x) = -x - 1/x. Now, -f(x) = -(x + 1/x) = -x - 1/x. Hey, these are equal! So, y = x + 1/x is indeed an odd function.But wait, I also need to check if it's increasing on (0, +∞). To determine if a function is increasing, I can look at its derivative. The derivative of y = x + 1/x is y' = 1 - 1/x². For the function to be increasing, y' should be positive for all x in (0, +∞).Let's see where y' is positive. 1 - 1/x² > 0 implies that 1 > 1/x², which simplifies to x² > 1, so x > 1 or x < -1. But since we're only considering x > 0, it means x > 1. So, y = x + 1/x is increasing on (1, +∞), but it's decreasing on (0, 1). Therefore, it's not increasing on the entire interval (0, +∞). So, option C doesn't satisfy both conditions.Finally, option D: y = x - 1/x.Let's check if it's odd. f(-x) = -x - 1/(-x) = -x + 1/x. On the other hand, -f(x) = -(x - 1/x) = -x + 1/x. These are equal, so y = x - 1/x is an odd function.Now, let's check if it's increasing on (0, +∞). Again, I'll find the derivative. The derivative of y = x - 1/x is y' = 1 + 1/x². Since x² is always positive for x ≠ 0, 1/x² is also positive. Therefore, y' = 1 + 1/x² is always positive for all x > 0. This means the function is increasing on the entire interval (0, +∞).So, option D satisfies both conditions: it's an odd function and it's increasing on (0, +∞).Just to recap:- A: Even function, not odd.- B: Not odd.- C: Odd but not increasing on the entire interval.- D: Odd and increasing on the entire interval.Therefore, the correct answer is D.

💡First, I need to determine the possible configurations of the isosceles triangle with side lengths of 2 and 4.In an isosceles triangle, two sides are equal. So, there are two possible cases:1. **Case 1:** The two equal sides are of length 2, and the base is 4.2. **Case 2:** The two equal sides are of length 4, and the base is 2.For each case, I will check if the triangle inequality holds, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side.**Case 1:**- Sides: 2, 2, 4- Check triangle inequality: - 2 + 2 = 4, which is not greater than 4. - Therefore, this configuration does not form a valid triangle.**Case 2:**- Sides: 4, 4, 2- Check triangle inequality: - 4 + 4 > 2 (8 > 2), which is true. - 4 + 2 > 4 (6 > 4), which is true. - 4 + 2 > 4 (6 > 4), which is true. - All conditions are satisfied, so this configuration forms a valid triangle.Now, I will calculate the perimeter of the valid triangle from Case 2:[text{Perimeter} = 4 + 4 + 2 = 10]

💡Alright, so I've got this function here: f(x) = |x - 3| + |x - 2| + k. There are two parts to the problem. Let me tackle them one by one.Starting with part (I): If f(x) is always greater than or equal to 3, find the range of values for k.Hmm, okay. So, f(x) is the sum of two absolute value expressions plus k. I know that absolute value functions have V-shapes, so their sum might have a minimum value somewhere. Maybe I need to find the minimum value of |x - 3| + |x - 2| and then set up an inequality involving k.Let me think. The expression |x - 3| + |x - 2| is the sum of distances from x to 3 and from x to 2 on the number line. I remember that the sum of distances from a point to two fixed points is minimized when the point is between the two fixed points. So, in this case, the minimum should occur somewhere between 2 and 3.Wait, actually, for two points, the sum of distances is minimized when x is between them. So, if x is between 2 and 3, the sum |x - 3| + |x - 2| would be equal to |3 - 2|, which is 1. So, the minimum value of |x - 3| + |x - 2| is 1.So, if I plug that back into f(x), the minimum value of f(x) would be 1 + k. Since f(x) has to be greater than or equal to 3 for all x, that means 1 + k must be greater than or equal to 3. So, solving for k, I get k >= 2.Wait, let me double-check that. If the minimum of |x - 3| + |x - 2| is 1, then f(x) = 1 + k must be >= 3. So, 1 + k >= 3 => k >= 2. Yeah, that seems right.Okay, so for part (I), k has to be greater than or equal to 2.Moving on to part (II): When k = 1, solve the inequality f(x) < 3x.So, substituting k = 1 into f(x), we get f(x) = |x - 3| + |x - 2| + 1. We need to find all x such that |x - 3| + |x - 2| + 1 < 3x.Hmm, dealing with absolute values can be tricky because they change their behavior depending on whether the expression inside is positive or negative. So, I think I need to break this down into cases based on the critical points where the expressions inside the absolute values change sign, which are at x = 2 and x = 3.So, the critical points are x = 2 and x = 3. Therefore, I can split the real number line into three intervals:1. x < 22. 2 <= x < 33. x >= 3For each interval, I'll rewrite the absolute value expressions without the absolute value signs, considering the sign in that interval, and then solve the inequality.Let's start with the first interval: x < 2.In this case, x - 3 is negative, so |x - 3| = -(x - 3) = 3 - x. Similarly, x - 2 is negative, so |x - 2| = -(x - 2) = 2 - x. So, substituting into f(x):f(x) = (3 - x) + (2 - x) + 1 = 3 - x + 2 - x + 1 = 6 - 2x.So, the inequality becomes 6 - 2x < 3x.Let me solve that:6 - 2x < 3xAdd 2x to both sides:6 < 5xDivide both sides by 5:6/5 < xSo, x > 6/5.But wait, in this interval, x < 2. So, the solution here is x > 6/5 and x < 2. So, the solution in this interval is (6/5, 2).Okay, moving on to the second interval: 2 <= x < 3.In this interval, x - 3 is still negative, so |x - 3| = 3 - x. However, x - 2 is non-negative, so |x - 2| = x - 2. So, f(x) becomes:f(x) = (3 - x) + (x - 2) + 1 = 3 - x + x - 2 + 1 = (3 - 2 + 1) + (-x + x) = 2 + 0 = 2.So, f(x) = 2 in this interval. The inequality becomes 2 < 3x.Solving for x:2 < 3x => x > 2/3.But in this interval, x is between 2 and 3. So, x > 2/3 is automatically satisfied because x >= 2. Therefore, the entire interval 2 <= x < 3 is a solution.Wait, hold on. If f(x) = 2 in this interval, then 2 < 3x implies x > 2/3. Since x is already >= 2, which is greater than 2/3, the inequality holds for all x in [2, 3). So, the solution here is [2, 3).Now, onto the third interval: x >= 3.In this case, both x - 3 and x - 2 are non-negative. So, |x - 3| = x - 3 and |x - 2| = x - 2. Therefore, f(x) becomes:f(x) = (x - 3) + (x - 2) + 1 = x - 3 + x - 2 + 1 = 2x - 4.So, the inequality is 2x - 4 < 3x.Let me solve that:2x - 4 < 3xSubtract 2x from both sides:-4 < xWhich is always true for x >= 3. So, in this interval, all x >= 3 satisfy the inequality.Therefore, combining the solutions from all three intervals:From the first interval: (6/5, 2)From the second interval: [2, 3)From the third interval: [3, ∞)So, putting it all together, the solution set is (6/5, ∞).Wait, hold on. Let me make sure I didn't make a mistake in combining the intervals. The first interval gives (6/5, 2), the second gives [2, 3), and the third gives [3, ∞). So, combining these, it's (6/5, ∞). Because from 6/5 to 2, then 2 to 3, and then 3 onwards, it's all connected. So, yeah, the solution is x > 6/5.But let me double-check the first interval. When x < 2, f(x) = 6 - 2x. So, 6 - 2x < 3x.6 < 5x => x > 6/5. But in this interval, x < 2, so x must be between 6/5 and 2. That's correct.In the second interval, f(x) = 2, so 2 < 3x => x > 2/3. But since x is already >= 2, which is greater than 2/3, so the entire interval [2, 3) is a solution.In the third interval, f(x) = 2x - 4 < 3x => x > -4, which is always true since x >= 3. So, all x >= 3 are solutions.Therefore, combining all the solutions, x must be greater than 6/5. So, the solution set is (6/5, ∞).Wait, but 6/5 is 1.2, right? So, x > 1.2. That seems reasonable.Let me just test a value in each interval to make sure.First interval: Let's pick x = 1.5, which is between 6/5 (1.2) and 2.f(x) = |1.5 - 3| + |1.5 - 2| + 1 = 1.5 + 0.5 + 1 = 3.So, 3 < 3*1.5 = 4.5. Yes, 3 < 4.5 is true.Another test in the first interval: x = 1, which is less than 6/5.f(x) = |1 - 3| + |1 - 2| + 1 = 2 + 1 + 1 = 4.4 < 3*1 = 3? No, 4 is not less than 3. So, x = 1 is not a solution, which is consistent with our solution set.In the second interval, let's pick x = 2.5.f(x) = |2.5 - 3| + |2.5 - 2| + 1 = 0.5 + 0.5 + 1 = 2.2 < 3*2.5 = 7.5. Yes, that's true.In the third interval, let's pick x = 4.f(x) = |4 - 3| + |4 - 2| + 1 = 1 + 2 + 1 = 4.4 < 3*4 = 12. Yes, that's true.What about x = 3?f(x) = |3 - 3| + |3 - 2| + 1 = 0 + 1 + 1 = 2.2 < 3*3 = 9. Yes, true.And x = 6/5 = 1.2.f(x) = |1.2 - 3| + |1.2 - 2| + 1 = 1.8 + 0.8 + 1 = 3.6.3.6 < 3*1.2 = 3.6? No, 3.6 is not less than 3.6. So, x = 6/5 is not included, which is why it's an open interval.Okay, so all the test points check out. So, I think my solution is correct.So, to recap:(I) The minimum value of |x - 3| + |x - 2| is 1, so 1 + k >= 3 => k >= 2.(II) When k = 1, solving f(x) < 3x gives x > 6/5, so the solution set is (6/5, ∞).**Final Answer**(I) The range of values for ( k ) is (boxed{k geqslant 2}).(II) The solution set of the inequality is (boxed{left( dfrac{6}{5}, +infty right)}).

💡Okay, so I have this problem about stratified sampling in a region with different types of schools. Let me try to understand what it's asking and figure out how to solve it step by step.First, the problem states there are 21 primary schools, 14 middle schools, and 7 universities. They want to use stratified sampling to select 6 schools for a vision survey. Then, part (2) asks about selecting 2 schools from these 6 for further analysis.Starting with part (1): I need to calculate how many primary, middle, and university schools should be selected. I remember that stratified sampling involves dividing the population into subgroups (strata) and then taking a proportional sample from each subgroup. So, the first thing I should do is find the total number of schools.Total schools = 21 primary + 14 middle + 7 universities = 42 schools.They want to select 6 schools out of 42. So, the sampling ratio would be 6/42, which simplifies to 1/7. That means we take 1/7 of each stratum.Calculating for primary schools: 21 * (1/7) = 3. So, 3 primary schools should be selected.Middle schools: 14 * (1/7) = 2. So, 2 middle schools.Universities: 7 * (1/7) = 1. So, 1 university.Wait, let me double-check that. 3 + 2 + 1 = 6, which matches the total number of schools to be selected. That seems correct.Moving on to part (2): Now, from these 6 selected schools, we need to randomly choose 2 for further data analysis.Part (i) asks to list all possible outcomes. So, we have 6 schools, and we need to list all possible pairs. Let me denote the primary schools as P1, P2, P3; middle schools as M1, M2; and the university as U1.So, the possible pairs are:1. P1 and P22. P1 and P33. P1 and M14. P1 and M25. P1 and U16. P2 and P37. P2 and M18. P2 and M29. P2 and U110. P3 and M111. P3 and M212. P3 and U113. M1 and M214. M1 and U115. M2 and U1So, there are 15 possible outcomes. That seems right because the number of ways to choose 2 out of 6 is 6C2 = 15.Part (ii) asks for the probability that both selected schools are primary schools. From the list above, the pairs that consist of both primary schools are:1. P1 and P22. P1 and P33. P2 and P3So, there are 3 favorable outcomes. The total number of possible outcomes is 15. Therefore, the probability is 3/15, which simplifies to 1/5.Let me just make sure I didn't miss any pairs or count incorrectly. There are 3 primary schools, so the number of ways to choose 2 is 3C2 = 3. And the total number of ways to choose any 2 from 6 is 15. So, 3/15 = 1/5 is correct.I think that covers everything. I don't see any mistakes in my calculations or reasoning.**Final Answer**(1) The number of primary, middle, and university schools selected are boxed{3}, boxed{2}, and boxed{1} respectively.(2) (ii) The probability that both selected schools are primary is boxed{dfrac{1}{5}}.

💡Okay, so I have this problem here about negating a proposition. The proposition is: "There exists an m₀ in the natural numbers N such that the square root of m₀ squared plus one is also a natural number." And I need to find the negation of this statement from the given options.First, let me make sure I understand what the original statement is saying. It's saying that there is at least one natural number m₀ where if you take m₀, square it, add one, and then take the square root of that result, you end up with another natural number. So, in mathematical terms, it's saying ∃m₀ ∈ N, √(m₀² + 1) ∈ N.Now, I need to find the negation of this statement. I remember that in logic, the negation of an existential quantifier (∃) is a universal quantifier (∀), and vice versa. So, the negation of "there exists" is "for all," and the negation of "for all" is "there exists."So, applying that to this problem, the negation of ∃m₀ ∈ N, √(m₀² + 1) ∈ N should be ∀m ∈ N, √(m² + 1) ∉ N. That makes sense because if it's not true that there exists such an m₀, then for every m in N, the square root of m squared plus one is not a natural number.Let me check the options given:A: ∀m ∉ N, √(m² + 1) ∈ N B: ∀m ∈ N, √(m² + 1) ∉ N C: ∃m₀ ∈ N, √(m₀² + 1) ∉ N D: ∃m₀ ∉ N, √(m₀² + 1) ∈ NOption A says that for all m not in N, the square root of m squared plus one is in N. That doesn't seem right because the original statement was about m being in N, so the negation shouldn't be about m not being in N.Option B says that for all m in N, the square root of m squared plus one is not in N. This aligns with what I was thinking earlier. If there doesn't exist an m₀ in N such that √(m₀² + 1) is in N, then for every m in N, √(m² + 1) must not be in N.Option C says that there exists an m₀ in N such that √(m₀² + 1) is not in N. That's actually not the negation of the original statement. The original statement was about existence, and the negation should be about universality, not another existence.Option D says that there exists an m₀ not in N such that √(m₀² + 1) is in N. Again, this is shifting the domain of m to not being in N, which doesn't seem to be the correct negation.So, based on this reasoning, Option B seems to be the correct negation of the original proposition.But wait, let me double-check. The original statement is ∃m₀ ∈ N, √(m₀² + 1) ∈ N. The negation should be that for all m in N, √(m² + 1) is not in N. So, yes, that's exactly what Option B is stating.Just to make sure, let's think about what this means in practical terms. If the original statement is true, there is at least one natural number m₀ where m₀² + 1 is a perfect square. For example, m₀ = 0: √(0 + 1) = 1, which is natural. Or m₀ = 1: √(1 + 1) = √2, which is not natural. Wait, so actually, m₀ = 0 would satisfy the original statement because √1 = 1 is natural. But if we consider m₀ starting from 1, then √(1 + 1) is not natural. Hmm, this might be a point of confusion.But regardless of specific examples, logically, the negation should follow the rules of quantifiers. The negation of "there exists" is "for all," and the negation of "is in N" is "is not in N." So, the structure should be ∀m ∈ N, √(m² + 1) ∉ N, which is Option B.I think I'm confident now that Option B is the correct answer.